hello :
x = 4 + ln t....(1)
y = t² + 3....(2)
1) the first method :
by (1) : lnt = x-4
t = e^(x-4)
subsct in (2) : y = (e^(x-4))²+3
y = e^(2x-8)+3
let : f(x) = e^(2x-8)+3
an equation of the tangent at (4,4) is : y = f'(4)(x-4) +f(4)
f'(x) = 2e^(2x-8)
f'(4) =2e^(0) =2
f(4) = 4
so : y = 2(x-4)+4
y= 2x-4
